# Finding the reason behind the value of the integral.

I was just trying to find $$\int_{0}^{\pi / 2}\frac{\sin{9x}}{\sin{x}}\,dx$$ using an online integral calculator. And surprisingly I found that if I replace $$9x$$ by $$x,3x,5x$$ which are some odd multiples of $$x$$ the value of integral came out to be $$\dfrac \pi 2$$.

I can't figure out the reason and would like to know why this is happening.

Edit: It can also be noted that $$\int_{a{\pi}}^{b\pi }\frac{\sin{9x}}{\sin{x}}\,dx =(b-a){\pi}$$ where $$a,b$$ are integers.

• The following identity seems like it may help:$$\frac{\sin((n+1/2)\theta}{\sin(\theta/2)}=1+2\cos x+2\cos(2x)+\cdots+2\cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.) – Semiclassical Apr 2 '19 at 14:46
• I think the statement about the integral from $a\pi$ to $b\pi$ is incorrect. Setting $a=\frac18$ and $b=\frac16,$ Wolfram Alpha says the integral is a negative number, not $(\frac16-\frac18)\pi.$ Did you mean to say instead that $a$ and $b$ are integers? – David K Apr 2 '19 at 21:12
• @David I checked for few other rational a and b and they satisfied but you are right. – Jasmine Apr 2 '19 at 21:34

## Hint

Consider $$I(n)=\int_{0}^{\pi/2} \frac{\sin(nx)}{\sin x} dx$$

$$I(2m+1)-I(2m-1)=\int_{0}^{\pi/2} \frac{\sin(2m+1)x-\sin(2m-1)x}{\sin{x}} dx=\int_{0}^{\pi/2} \frac{2\sin(x)\cos(2mx)}{\sin{x}} dx$$ $$\implies 2\int_{0}^{\pi/2} \cos(2mx)dx.......(1)$$ Now think what happens to this integral when $$m$$ is an integer. And also try to use the fact $$I(1)=\frac{\pi}{2}$$.

Edit (As OP has changed the question a bit)

Now consider$$I(n)=\int_{a\pi}^{b\pi} \frac{\sin(nx)}{\sin x} dx$$

From(1) $$\implies 2\int_{a\pi}^{b\pi} \cos(2mx)dx=2\bigg[\frac{\sin(2mx)}{2m}\bigg]_{a\pi}^{b\pi}$$ $$\implies I(2m+1)-I(2m-1)=\frac{1}{n} \bigg[\sin(2\pi bx)-\sin(2\pi ax)\bigg]=0$$ Provided $${a,b} \in \mathbb{Z}$$ $$\implies I(2m+1)=I(2m-1)$$ Now As $$I(1)=(b-a)\pi$$

Hence

$$\bbox[5px,border:2px solid blue] {\int_{a\pi}^{b\pi} \frac{\sin(nx)}{\sin x} dx=(b-a)\pi }$$ When n is odd.